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\title{\vspace{-4cm}\textbf{河北师范大学数学分析真题}}
\author{宁鑫雨}
\date{\today}
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\begin{document}
\date{}
\section*{2010年数学分析}
\begin{problem}[本题30分]
     
    1.求极限 $\displaystyle\lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} \right)$\\
    2.设 $y=y(x)$可微，且满足 $y=-ye^x+2e^y\sin x-7x$,求$y'(0)$.\\
    3.求球面 $x^2+y^2+z^2=a^2.(a>0)$被平面 $z=\frac{a}{4}$与 $z=\frac{a}{2}$所夹部分的曲面面积.
\end{problem}
\begin{solution}[本题30分,每题10分]\\
    1)\\
    $\begin{aligned}
        &\lim_{n\to\infty}(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}) \\
        &=\lim_{n\to\infty}\frac{1}{n}(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdots+\frac{1}{1+\frac{n}{n}}) \\
        &=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}} \\
        &=\int_{0}^{1}\frac{1}{1+x}\d x \\
        &=\ln|1+x|\big|_{0}^{1} \\
        &=\1n2
    \end{aligned}$\\
    2)将$x=0$代入$y(x)=-y(x)e^{x}+2e^{y(x)}\sin x-7x$\\
    得$y(0)=-y(0)\Longrightarrow y(0)=0$\\
    将$y(x)=-y(x)e^{x}+2e^{y(x)}\sin x-7x$两边对$x$求导\\
    $y^{\prime}(x)=-y^{\prime}(x)e^{x}-y(x)e^{x}+2e^{y(x)}\cdot y^{\prime}(x)\sin x+2e^{y(x)}\cos x-7$\\
    $\begin{aligned}
        \therefore y^{\prime}(0)
        &=-y^{\prime}(0)e^{0}-y(0)e^{0}+0+2e^{y(0)}\cdot1-7\\
        &=-y^{\prime}(0)-5
    \end{aligned}$\\
    $\therefore y^{\prime}(0)=-\frac{5}{2}$\\
    3)$z=\sqrt{a^{2}-x^{2}-y^{2}}$\\
    $S=\iint\sqrt{1+(z_{x})^{2}+(z_{y})^{2}}\d x\d y$\\
    $z_{x}=\frac{x}{\sqrt{a^{2}-x^{2}-y^{2}}},z_{y}=\frac{y}{\sqrt{a^{2}-x^{2}-y^{2}}}$\\
    $\sqrt{1+z_{x^{2}}+z_y^{2}}=\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}}$\\
    $\therefore S=\iint_{D_{xy}}\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}}\d x\d y ,D_{xy}:\frac{\sqrt{3}a}{2}\leq\sqrt{x^{2}+y^{2}}\leq\frac{\sqrt{15}a}{4}$\\
    令$x=r\cos\theta,y=r\sin\theta,0\leq\theta\leq2\pi$\\
    $\frac{\sqrt{3}a}{2}\leq r \leq\frac{\sqrt{15}a}{4}$\\
    $\begin{aligned}
        \therefore S&=\iint_{Dxy}\frac{a}{\sqrt{a^{2}-x^{2}+y^{2}}}\d x\d y\\
        &=\int_{0}^{2\pi}x\ d \theta\int_{\frac{\sqrt{3}a}{2}}^{\frac{\sqrt{15}a}{4}}\frac{a\cdot r}{\sqrt{a^{2}-r^{2}}}\ d r\\
        &=\frac{a^{2}\pi}{2}
    \end{aligned}$
\end{solution}

\begin{problem}[本题15分]
    设$\left\{ a_n \right\}$是数列\\
    1. 若$\left\{ a_n \right\}$ 单调递增，且有一个子列$\left\{ a_{n_i} \right\}$ 收敛，证明$\left\{ a_n \right\}$收敛.\\
    2.若数列$\left\{ a_n \right\}$的两个子列 $\left\{ a_{2n} \right\}$与$\left\{ a_{2n-1} \right\}$都收敛，且极限相同，证明$\left\{ a_n \right\}$收敛.
\end{problem}
\begin{solution}[本题15分]\\
    1)$\{a_{n_k}\}$收敛,设收敛于$a$\\
    $\therefore$对于$\varepsilon=1,\exists N$,当$k>N$时,$|a_{n_k}-a|<1$\\
    $\therefore a_{n_{k}}<a+1=M,$则$M$也是$\{a_{n}\}$的上界\\
    若不然，存在$N$,当$a_{N}>M,$又$\because\{a_{n}\}$递增\\
    $\therefore$当$n>N$时,$a_n>M.$\\
    $\therefore$当$k>N$时,$a_{n_k}\geq a_{k}>M$与$M$为$\{a_{n}\}$的上界矛盾\\
    $\therefore \{a_{n}\}$单增有上界$\therefore$由单调有界定理极限存在\\
    2)设$\{a_{2n}\},\{a_{2n-1}\}$收敛于0\\
    则对$\forall \varepsilon>0$,在$U(x;\varepsilon)$之外有$\{a_{2n}\},\{a_{2n-1}\}$的有限项,\\
    $\therefore$在$U(x;\varepsilon)$之外有$\{a_{n}\}$的有限项,\\
    $\therefore \{a_{n}\}$收敛,且也收敛于0
\end{solution}

\begin{problem}[本题15分]
    设$f(x)$在$[-1,1]$上有二阶导数，且  $f(-1)=f(1)=\displaystyle\frac{1}{2}$,$\displaystyle|f''(x)|\le \frac{1}{2}$ 证明：\\
    1.$|f'(x)|\le \displaystyle\frac{1}{2},x\in \left[ -1,1 \right]$.\\
    2.$f(x)=x$在 $[-1,1]$ 上有且仅有一个实数根. 
\end{problem}
\begin{solution}[本题15分]\\
   证明:\\
   1)$f(-1)=f(x)+f^{\prime}(x)(-1-x)+\frac{f^{\prime\prime}(\xi_{1})}{2}(-1-x)^{2},-1<\xi_{1}<x,\textcircled{1}$\\
   $f(1)=f(x)+f^{\prime}(x)(1-x)+\frac{f^{\prime\prime}(\xi_{2})}{2}(1-x)^{2},x<\xi_{2}<1,\textcircled{2}$\\
   $\textcircled{1}-\textcircled{2}$得$2f^{\prime}(x)=\frac{(1+x)^{2}}{2}f^{\prime\prime}(\xi_{1})-\frac{(1-x)^{2}}{2}f^{\prime\prime}(\xi_{2})$\\
   $\begin{aligned}
    \therefore2\left|f^{\prime}(x)\right|
    &=\left|\frac{f(\xi_1)}{2}\left(1+x\right)^2-\frac{f^{\prime\prime}(\xi_2)}{2}\left(1-x\right)^2\right|\\
    &=\left|f^{\prime\prime}(\xi_1)\right|\frac{\left(1+x\right)^2}{2}+\left|f^{\prime\prime}(\xi_2)\right|\frac{\left(1-x\right)^2}{2}\\
    &\leqslant\frac{(1+x)^2+(1-x)^2}4
    \end{aligned}$\\
    $\therefore|f^{\prime}(x)|\leq\frac{(1+x)^{2}+(1-x)^{2}}{8},x\in[1,1]$\\
    $\frac{(1+x)^{2}+(1-x)^{2}}{8}=\frac{x^{2}+1}{4}\leq\frac{1}{2}$\\
    $\therefore|f^{\prime}(x)|\leq\frac{1}{2}\cdot,x\in[-1,1]$\\
    2)令$F(x)=f(x)-x$\\
    $F(-1)=f(-1)+1=\frac{3}{2},F(1)=f(1)-1=\frac{1}{2}$\\
    又$\because f(x)$上有二阶导数\\
    $\therefore f(x)$连续,$\therefore F(x)$连续\\
    $\therefore$由介值定理,$F(x)$至少有一个零点,$x \in [-1,1]$\\
    $F^{\prime}(x)=f^{\prime}(x)-1$\\
    由1知,$F^{\prime}(x)<0$,$\therefore F(x)$单减\\
    $\therefore F(x)$在$[-1,1]$上至少只有一个零点\\
    $\therefore F(x)$在$[-1,1]$上有且仅有一个零点,即$f(x)=x$有且仅有一个实根
\end{solution}

\begin{problem}[本题15分]
    设 $f(x)$ 在任何有限区间上可积，且 $\displaystyle\lim_{x \to \infty}f(x)=l<+\infty $,证明：
    $\displaystyle\lim_{x\to +\infty }\frac{1}{x}\int_0^xf(t)\d t=l$.  
\end{problem}
\begin{solution}[本题15分]\\
   证明:\\
   $\because\lim_{x\to+\infty}f(x)=l$\\
   $\therefore \forall \varepsilon>0,\exists N>0,$当$x>N$时,$l-\varepsilon<f(x)<l+\varepsilon$\\
   $\frac{1}{x}\int_{0}^{x}f(t)\ d t=\frac{1}{x}\int_{0}^{N}f(t)\ d t+\frac{1}{x}\int_{N}^{x}f(t)\ d t$\\
   $\lim_{x\rightarrow+\infty}\frac{1}{x}\int_{0}^{N}f(t)\ d t=0$\\
   又$\frac{1}{x}\int_{N}^{x}(l-\varepsilon)\ d t<\frac{1}{x}\int_{N}^{x}f(t)\ d t<\frac{1}{x}\int_{N}^{x}(l+\varepsilon)\ d t$\\
   $-\varepsilon(\frac{x-N}{x})<\frac{1}{x}\int_{N}^{x}f(t)\ d t-l(\frac{x-N}{x})<\varepsilon(\frac{x-N}{x})$\\
   $\therefore|\frac{1}{x}\int_{N}^{x}f(t)\ d t-l|=|\frac{1}{x}\int_{N}^{x}(f(t)-l)\ d t+\frac{x-N}{x}l-l|<\varepsilon \cdot\frac{x-N}{x}+\frac{Nl}{x}$\\
   $x\rightarrow\infty,\frac{x-N}{x}\rightarrow1,\frac{Nl}{x}\rightarrow0$\\
   $\therefore|\frac{1}{x}\int_{N}^{x}f(t)\ d t-l|<\varepsilon$\\
   $\therefore\lim_{x\to+\infty}\frac{1}{x}\int_{N}^{x}f(t)\ d t=l$\\
   $\therefore\lim_{x\to0}\frac{1}{x}\int_{0}^{x}f(t)\ d t=\lim_{x\to0}\frac{1}{x}\int_{0}^{N}f(t)\ d t+\lim_{x\to0}\frac{1}{x}\int_{N}^{x}f(t)\ d t=l$
\end{solution}
    
\begin{problem}[本题15分]
    设 $f(x,y)$ 是 $D=[a,b]\times [c,d]$ 上的连续函数，$\left\{ \varphi_n(t) \right\},\left\{ \psi_n(t) \right\}$ 在 $[\alpha,\beta]$
    上一致收敛，且 $a\le \varphi_n(t)\le b$ ,$c\le \psi_n(t) \le d$ ,令 $\displaystyle x_n=\int_\alpha^\beta f\left( \varphi_n(t),\psi_n(t) \right)\d t$\\ 
    1.证明数列 $\left\{ x_n \right\}$ 收敛.\\ 
    2.如果把一致收敛的条件改为收敛，结论是否成立？
\end{problem}
\begin{solution}[本题15分]\\
  证明:\\
  $\because  \{{\varphi}_{n}(t)\},\{{\psi}_{n}(t)\}$在$[\alpha,\beta]$上一致收敛\\
  并设为分别收敛于$\varphi(t),\psi(t)$\\
  $\therefore$对$\forall \varepsilon>0,\exists N,$\\
  当$n>N$时,$|{\varphi}_{n}(t)-\varphi(t)|<\varepsilon,|{\psi}_{n}(t)-\psi(t)|<\varepsilon,t \in [\alpha,\beta]$\\
  又$\because f(x,y)$在$D$上连续\\
  $\therefore f(x,y)$在$D$上一致连续\\
  $\therefore \forall\varepsilon^{\prime}>0,\exists\delta=\varepsilon,$\\
  当$|x_{1}-x_{2}|<\delta=\varepsilon,|y_{1}-y_{2}|<\delta=\varepsilon $,$|f(x,y_{1})-f(x_{1},y_{2})|<\frac{\varepsilon}{\beta-\alpha}$\\
  $\therefore$有$|{\varphi}_{n}(t){\psi}_{n}(t)-\varphi(t)\psi(t)|<\frac{\varepsilon}{\beta-\alpha}$\\
  $\begin{aligned}
    &\big|\int_{\alpha}^{\beta}f({\varphi}_{n}(t),\psi_{n}(t))\ d t-\int_{\alpha}^{\beta}f({\varphi}(t),\psi(t))\ d t\big|\\
    &\leq\int_{\alpha}^{\beta}\big|f({\varphi}_{n}(t),\psi_{n}(t))\ d t-f({\varphi}(t),\psi(t))\ d t\big|\\
    &<\frac{\varepsilon^{\prime}(\beta-\alpha)}{\beta-\alpha}=\varepsilon^{\prime}
    \end{aligned}$\\
    $\therefore\lim_{n\to\infty}x_{n}=\int_{\alpha}^{\beta}f({\varphi}(t),\psi(t))\ d t$\\
    即${x_{n}}$收敛\\
    2)不一定成立,无法找到$N(\varepsilon)$
  
\end{solution}
    
\begin{problem}[本题15分]
    设 $f(x)$ 在点 $x=1$ 具有二阶导数，且 $f''(1)=1$ ，$\displaystyle\lim_{x\to 1}\frac{f(x)}{x-1}=0$ .令
    $\displaystyle g(x)=\int_0^1f'\left[ 1+(x-1)t \right]\d t$ .求 $g'(1)$ .
    
\end{problem}
\begin{solution}[本题15分]\\
   证明:\\
   容易得出$f(1)=0,f^{\prime}(1)=0$\\
   令$u=1+(x-1)t$\\
   $\therefore g(x)=\int_{0}^{1}f^{\prime}[1+(x-1)t]\ d t=\frac{1}{x-1}\int_{1}^{x}f^{\prime}(u)\ d u$\\
   易知$g(1)=0$\\
   $\begin{aligned}
    &g^{\prime}(u)=\lim_{x\to1}\frac{g(x)-g(1)}{x-1}\\
    &=\lim_{x\to1}\frac{\int_{1}^{x}f^{\prime}(u)\ d u}{(x-1)^{2}}\\
    &=\lim_{x\to1}\frac{f^{\prime}(x)}{2(x-1)}=\lim_{x\to1}\frac{f^{\prime\prime}(x)}{2}=\frac{1}{2}
    \end{aligned}$\\
    $\therefore g^{\prime}(1)=\frac{1}{2}$
\end{solution}

\begin{problem}[本题15分]
    设 $f(x)=\displaystyle\sum_{n=1}^{+\infty }\frac{\sin{nx}}{\sqrt{n^5+4}}$,证明： \\
    1.$f(x)$ 在 $\left( -\infty ,+\infty  \right)$ 可导.\\
    2.$\displaystyle-\frac{1}{\sqrt{5}}\le f'(\pi) \le -\frac{1}{\sqrt{5}}+\frac{1}{3}$ 
\end{problem}
\begin{solution}[本题15分]
    证明:\\
    1)$|\frac{\sin x}{\sqrt{n^{5}+\varphi}}|\leq\frac{1}{\sqrt{n^{5}}},\forall x \in(-\infty,+\infty)$\\
    $\because\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^{5}}}$收敛\\
    $\therefore$由优级数判别法知$\sum_{n=1}^{\infty}\frac{\sin x}{\sqrt{n^{5}+\varphi}}$一致收敛\\
    $\sum_{n=1}^{\infty}\frac{\sin x}{\sqrt{n^{5}+\varphi}}$收敛\\
    又每一项都可导$(\frac{\sin x}{\sqrt{n^5+\varphi}})^{\prime}=\frac{n\cos nx}{\sqrt{n^5+\varphi}}$\\
    且$|\frac{1}{\sqrt{n^{5}+\varphi}}|\leq\frac{1}{\sqrt{n^{3}}}$\\
    $\therefore$由优级数判别法知$\sum_{n=1}^{\infty}\frac{n\cos nx}{\sqrt{n^{5}+\varphi}}$一致收敛\\
    $f(x)$在$(-\infty,+\infty)$可导,\\
    $\therefore f^{\prime}(x)=\sum_{n=1}^{+\infty}\frac{n\cos nx}{\sqrt{n^{5}+\varphi}}$\\
    2)\\
    $\begin{aligned}
        &f^{\prime}(\pi)=\sum_{n=1}^{\infty}\frac{n\cos n\cdot\pi}{\sqrt{n^{5}+\varphi}} \\
        &=-\frac{1}{\sqrt{5}}+\frac{1}{3}-a_{1}+a_{2}-a_{3}+\cdots  \\
        &=-\frac{1}{\sqrt{5}}+\frac{1}{3}-(a_{1}-a_{2})-(a_{3}-a_{4})-\cdots  \\
        &=-\frac{1}{\sqrt{5}}+(\frac{1}{3}-a_{1})+(a_{2}-a_{3})+\cdots 
    \end{aligned}$\\
    $\{\frac{n}{\sqrt{n^{5}+\varphi}}\}$单减趋于0,\\
    $\therefore$由莱布尼茨知$f^{\prime}(\pi)$收敛\\
    $\therefore-\frac{1}{\sqrt{5}}<f^{\prime}(\pi)<-\frac{1}{\sqrt{5}}+\frac{1}{3}.$
\end{solution}

\begin{problem}[本题15分]
    设 $f(x)$ 在 $[a,b]$ 上连续，证明：
    $$ 2\int_a^b \left( f(x) \int_x^bf(t)\d t  \right) \d x = \left( \int_a^b f(x) \d x\right)^2$$
\end{problem}
\begin{solution}[本题15分]
   证明:\\
   设$\int_{x}^{b}f(t)\ d t=g(x),g^{\prime}(x)=-f(x)$\\
   $\begin{array}{l}
    {\therefore2\int_{a}^{b}\{f(x)\int_{x}^{b}f(x)\ d t\}\ d x=-2\int_{a}^{b}g^{\prime}(x)g(x)\ d x}\\
    {=-2\int_{a}^{b}g(x)\ d g(x)=-g^{2}(x)\big|_{a}^{b}=g^{2}(a)-g^{2}(b)}\\
    \end{array}$\\
    $g(a)=\int_{a}^{b}f(t)\ d t,g(b)=0$\\
    $\therefore2\int_{a}^{b}\{f(x)\int_{x}^{b}f(t)\ d t\}\ d x=(\int_{a}^{b}f(t)\ d t)^{2}$
\end{solution}

\begin{problem}[本题15分]
    设函数 $f(x)$ 在 $[a,b]$ 有定义\\
    1.若 $x_0 \in [a,b]$ 是 $f(x)$ 的第一类间断点，证明：存在 $x_0$ 的邻域 $U(x_0)$ 使得 $f(x)$ 在 $U(x_0)\bigcap[a,b]$ 有界.\\ 
    2.若 $f(x)$ 在 $[a,b]$ 只有第一类间断点，问 $f(x)$ 在 $[a,b]$ 是否一定有界？验证你的结论.
\end{problem}
\begin{solution}[本题15分]
   证明:\\
   1)$\because x_{0}$为$f(x)$的第一类间断点\\
   $\therefore$存在$x_{0}$的左邻域和右邻域\\
   $\therefore f(x)$在这两个邻域内均有界\\
   但$f(x)$在$x_{0}$有定义\\
   $\therefore $存在$x_{0}$的邻域$U(x_{0})$,使得$f(x)$在$U(x_{0}) \cap [a,b]$有界\\
   同理,当$x_{0}$为$a$或$b$时,结论同样成立\\
   2)$\forall x \in [a,b]$,$x$为$f(x)$的第一类间断点\\
   则存在$U(x)$,使得$f(x)$在$U(x) \cap [a,b]$上有界\\
   而$\{U(x):x\in[a,b]\}$是$[a,b]$的一个覆盖\\
   由有限覆盖定理,存在$\{U(x):x\in[a,b]\}$的一个有限子集$\{U(x_{i}):i=1,2\cdots k\}$覆盖了$[a,b]$\\
   且$f(x)$在每一个$U(x_{i})\cap[a,b]$有界\\
   $\therefore f(x)$在$[a,b]$有界
\end{solution}
\end{document}